Question

Which of the following equations is of a parabola with a vertex at (0, -5)?
1. y = x2 - 5
2. y = x2 + 5
3. y = ( x + 5) 2
4. y = ( x - 5) 2

Answer 1

Answer:

The equation that has a vertex in (0, -5) is the first one.

Step-by-step explanation:

A standard form second degree equation is given by the following expression:

$y(x) = a*x^2 + b*x + c$

For which the vertex coordinates can be calculated by:

$x_{vertex} = -\frac{b}{2*a}$

While "y" for the vertex can be found by applying this coordinate on the expression. Using this knowledge in each equation gives us:

1. $y(x) = x^2 - 5$

$a = 1\\b = 0\\c = -5$

Therefore the vertex coordinate is:

$x_{vertex} = -\frac{b}{2*a} = -\frac{0}{2*1} = 0$

$y_{vertex} = 0^2 - 5 = -5$

This parabola has a vertex in (0,-5).

2. $y(x) = x^2 + 5$

$a = 1\\b = 0\\c = 5$

Therefore the vertex coordinate is:

$x_{vertex} = -\frac{b}{2*a} = -\frac{0}{2*1} = 0$

$y_{vertex} = 0^2 + 5 = 5$

This parabola has a vertex in (0,5).

3. $y(x) =(x+ 5)^2 = x^2 + 10*x + 25$

$a = 1\\b = 10\\c = 25$

Therefore the vertex coordinate is:

$x_{vertex} = -\frac{b}{2*a} = -\frac{10}{2*1} = -5$

$y_{vertex} = (-5)^2 + 10*(-5) + 25 = 0$

This parabola has a vertex in (-5,0).

4. $y(x) =(x- 5)^2 = x^2 - 10*x + 25$

$a = 1\\b =- 10\\c = 25$

Therefore the vertex coordinate is:

$x_{vertex} = -\frac{b}{2*a} = -\frac{-10}{2*1} = 5$

$y_{vertex} = (5)^2 - 10*(5) + 25 = 0$

This parabola has a vertex in (5,0).